Java 比较多个字符串

Java 比较多个字符串,java,android,string-comparison,Java,Android,String Comparison,我要做的是检查多个sring是否为null,以便转到下一个活动,类似这样: if( string1.equal(null) && ... stringN.equal(null)) { Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();} else { GO TO THE NEXT ACTIVITY} private String[] array; private String one =

我要做的是检查多个sring是否为null,以便转到下一个活动,类似这样:

if( string1.equal(null) && ... stringN.equal(null))
{ Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();}
else
{ GO TO THE NEXT ACTIVITY}
private String[] array;

private String one = "one";
private String two = "two";
private String three = "three";
private String four = "four";
private String five = "five";


private void someMethod(){
    if(isStringsNotNull(getStringArray())){
        //Start another activity
    } else {
        //Do something
    }
}

private String[] getStringArray(){
    return new String[]{one, two, three, four, five};
}

private Boolean isStringsNotNull(String[] array){
    for(String str: array){
        if(str == null){
            return false;
        }
    }
    return true;
}

所有字符串都必须具有内容。。。如果字符串位于列表或数组中,则任何帮助都将非常有用,然后可以迭代它们

如果使用ArrayList:

private boolean verify(final ArrayList<String> list)
{
    for (String entry : list)
    {
        if (TextUtils.isEmpty(entry))
            return false;
    }

    return true;
}

如果字符串位于列表或数组中,则可以迭代它们

如果使用ArrayList:

private boolean verify(final ArrayList<String> list)
{
    for (String entry : list)
    {
        if (TextUtils.isEmpty(entry))
            return false;
    }

    return true;
}

您也可以这样做:

if( string1.equal(null) && ... stringN.equal(null))
{ Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();}
else
{ GO TO THE NEXT ACTIVITY}
private String[] array;

private String one = "one";
private String two = "two";
private String three = "three";
private String four = "four";
private String five = "five";


private void someMethod(){
    if(isStringsNotNull(getStringArray())){
        //Start another activity
    } else {
        //Do something
    }
}

private String[] getStringArray(){
    return new String[]{one, two, three, four, five};
}

private Boolean isStringsNotNull(String[] array){
    for(String str: array){
        if(str == null){
            return false;
        }
    }
    return true;
}

回答得好@PerracoLabs,但我只能检查长度是否为0。不是吗?

你也可以这样做:

if( string1.equal(null) && ... stringN.equal(null))
{ Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();}
else
{ GO TO THE NEXT ACTIVITY}
private String[] array;

private String one = "one";
private String two = "two";
private String three = "three";
private String four = "four";
private String five = "five";


private void someMethod(){
    if(isStringsNotNull(getStringArray())){
        //Start another activity
    } else {
        //Do something
    }
}

private String[] getStringArray(){
    return new String[]{one, two, three, four, five};
}

private Boolean isStringsNotNull(String[] array){
    for(String str: array){
        if(str == null){
            return false;
        }
    }
    return true;
}

回答得好@PerracoLabs,但我只能检查长度是否为0。不是吗?

如果您使用的是java8/lambda,您可以尝试使用以下函数:

List<String> stringList = new ArrayList<>(Arrays.asList(string2.split(" "))); // Split the string into list
        // compare the list with the source of the string, if match put into the list of result
        List<String> result = stringList.stream().filter(new Predicate<String>() {
            @Override
            public boolean test(String s) {
//                return string1.contains(s); // case sensitive
                return Pattern.compile(Pattern.quote(s), Pattern.CASE_INSENSITIVE).matcher(string1).find(); // case insensitive
            }
        }).collect(Collectors.toList());

        // check the match results
        if(result.size()<stringList.size()){
            Log.d("test", "onCreate: not match");
        }else{
            Log.d("test", "onCreate: match");
        }
List-stringList=newarraylist(Arrays.asList(string2.split(“”));//将字符串拆分为列表
//如果将匹配项放入结果列表中,请将列表与字符串源进行比较
List result=stringList.stream().filter(新谓词(){
@凌驾
公共布尔测试(字符串s){
//返回string1.contains;//区分大小写
返回Pattern.compile(Pattern.quote,Pattern.CASE不区分大小写)。matcher(string1.find();//不区分大小写
}
}).collect(Collectors.toList());
//检查比赛结果

如果(result.size()如果您使用的是java8/lambda,则可以尝试使用以下函数:

List<String> stringList = new ArrayList<>(Arrays.asList(string2.split(" "))); // Split the string into list
        // compare the list with the source of the string, if match put into the list of result
        List<String> result = stringList.stream().filter(new Predicate<String>() {
            @Override
            public boolean test(String s) {
//                return string1.contains(s); // case sensitive
                return Pattern.compile(Pattern.quote(s), Pattern.CASE_INSENSITIVE).matcher(string1).find(); // case insensitive
            }
        }).collect(Collectors.toList());

        // check the match results
        if(result.size()<stringList.size()){
            Log.d("test", "onCreate: not match");
        }else{
            Log.d("test", "onCreate: match");
        }
List-stringList=new-ArrayList(Arrays.asList(string2.split(“”));//将字符串拆分为列表
//如果将匹配项放入结果列表中,请将列表与字符串源进行比较
List result=stringList.stream().filter(新谓词(){
@凌驾
公共布尔测试(字符串s){
//返回string1.contains;//区分大小写
返回Pattern.compile(Pattern.quote,Pattern.CASE不区分大小写)。matcher(string1.find();//不区分大小写
}
}).collect(Collectors.toList());
//检查比赛结果

如果(result.size()如果要检查的字符串数量可变,我将执行以下操作:

private boolean validate(String... items) {
  for (String item : items) {
    if (TextUtils.isEmpty(item)) {
      return false;
    }
  }
  return true;
}
然后,您的代码将使用验证所需的字符串调用validate方法:

if (validate(string1, string2, stringN)) {
  Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();
} else { 
  // GO TO THE NEXT ACTIVITY
}

如果要检查的字符串数量可变,我将执行以下操作:

private boolean validate(String... items) {
  for (String item : items) {
    if (TextUtils.isEmpty(item)) {
      return false;
    }
  }
  return true;
}
然后,您的代码将使用验证所需的字符串调用validate方法:

if (validate(string1, string2, stringN)) {
  Toast.makeText(this, "something is empty", Toast.LENGTH_SHORT).show();
} else { 
  // GO TO THE NEXT ACTIVITY
}